A projectile of mass 200 g is launched in a viscous medium at an angle 60° with the horizontal, with an initial velocity of 270 m/s. It experiences a viscous drag force \(\vec{F}=-c \vec{v}\) where the drag coefficient \(c=0.1\) kg/s and \(\vec{v}\) is the instantaneous velocity of the projectile. The projectile hits a vertical wall after 2 s. Taking \(e=2.7\), the horizontal distance of the wall from the point of projection (in m) is

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Laws of Motion

A projectile of mass 200 g is launched in a viscous medium at an angle 60° with the horizontal, with an initial velocity of 270 m/s. It experiences a viscous drag force

F = - c v

where the drag coefficient

c = 0.1

kg/s and

v

is the instantaneous velocity of the projectile. The projectile hits a vertical wall after 2 s. Taking

e = 2.7

, the horizontal distance of the wall from the point of projection (in m) is

Answer: 170

Solution

In the horizontal direction (no gravity component):

m \frac{d v _{x}}{d t} = - c v_{x}

Solving:

v_{x} = v_{0 x} e^{- c t / m} = 270 cos 60° \cdot e^{- 0.1 t /0.2} = 135 e^{- t /2}

Horizontal distance:

S_{x} = \int_{0}^{2} v_{x} d t = 135 \int_{0}^{2} e^{- t /2} d t = 135 \times 2 [1 - e^{- 1}]

S_{x} = 270 (1 - \frac{1}{2.7}) = 270 \times \frac{1.7}{2.7} = 170 m

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer

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JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Laws of Motion

Solution

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